# Proof: Why the Root Mean Square of two positive numbers is always greater than their Geometric Mean?

This article focuses on discussing the proof of why the root-mean-square of two positive numbers is always greater than their geometric mean. Before getting into the details of the proof, let’s first discuss the basic terminologies:

**Root Mean Square (RMS)****:**

Consider some numbers, A_{1}, A_{2}, A_{3}, A_{4}, …. A_{N}, the root-mean-square of these numbers will be equal to the square root of the arithmetic mean of the squares of these numbers, i.e.

For A_{1}, A_{2}, A_{3}... A_{N}, the RMS value equals to-Root Mean Square (RMS) =√(((A_{1})^{2}+ (A_{2})^{2}+ (A_{3})^{2}+...+ (A_{N})^{2})/N) For two numbers, A and B, their RMS value equals to-Root Mean Square (RMS) =√((A^{2}+ B^{2})/2)

RMS is also known as the Quadratic mean and has a very wide range of applications. In statistics, it is often used as an alternative to the term, standard deviation. It is also used in many concepts related to physics, like electricity, etc.

**Geometric Mean (GM)****:**

Consider some numbers, A_{1}, A_{2}, A_{3}, A_{4}, …. A_{N}, the geometric mean of these numbers will be equal to the nth square root of the multiplication of all these numbers, i.e.

For A_{1}, A_{2}, A_{3}... A_{N}, the GM value equals to-Geometric Mean (GM) =^{n}√(A_{1}* A_{2}* A_{3}* ... * A_{N}) For two numbers, A and B, the GM value equals to-Geometric Mean (GM) =√(A * B)

It is used to find the average or mean of geometric sequences, which helps us in studying and analyzing many real-life concepts, like the growth of bacteria, investments, etc.

**Problem Statement:**

The root-mean-square of two positive numbers is always greater than the geometric mean for the same two numbers.

**Solution:**

Consider two numbers, A and B, such that A, B > 0 and A ≠ B.

1. Consider the formula of the square of the difference between two positive numbers-

(A - B)^{2}= A^{2}- 2AB + B^{2}--- (1)

2. It is known that the square of a number is always greater than or equal to zero. But here we have A ≠ B. So, it can be written as-

(A - B)^{2}> 0

3. Expanding the left-hand side of the above equation using the property defined in equation 1-

A^{2}- 2AB + B^{2}> 0

4. After doing some rearrangements-

A^{2}+ B^{2}> 2AB

5. Diving the above equation by 2-

((A^{2}+ B^{2})/2) > (A * B)

6. Since both the sides are greater than 0, i.e. positive. After squaring both sides, the following equation is obtained-

√((A√^{2 + B2)/2) > }(A * B)

Notice that the **left-hand side is the root-mean-square and the right-hand side is equal to the geometric mean of A and B**. This proves that the root-mean-square is always greater than the geometric mean for two positive numbers.